4  Quotient Groups

We start this chapter with three examples that are intended to motivate much of what follows.

Example 4.1 Consider the group \(\Z_6\) and one of its subgroups \(H = \{0, 3\}\). The distinct left cosets of \(H\) in \(\Z_6\) are: \[eH = 0 + \{0, 3 \} = \{ 0, 3 \} = 3 + \{0, 3 \} = 3H,\] \[1H = 1 + \{ 0, 3 \} = \{ 1, 4 \} = 4 + \{ 0, 3 \} = 4H,\] \[2H = 2 + \{ 0, 3 \} = \{ 2, 5 \} = 5 + \{ 0, 3 \} = 5H.\] We then re-order the row and column of the Cayley table for \(\Z_6\) in ‘coset order’.

On replacing ‘blocks’ by cosets we obtain:

\[\begin{array}{c|ccccc} \circ & H && 1+ H && 2+H \\ \hline\\ H & H && 1+H && 2+H \\\\ 1+H & 1+H && 2+H && H \\\\ 2+H & 2+H && H && 1+H \end{array}\]

This looks like a Cayley table for a group of cosets - but note that we do not have a binary operation.

Finally, if we ignore the \(H\)’s and use the fact that \(0+H = H\), we obtain the following:

\[\begin{array}{c|ccc} \circ & 0 & 1 & 2\\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \end{array}\]

So, it would appear that if \(H\) is a subgroup of \(G\), then we can ‘divide’ \(G\) by \(H\) to obtain a group of order \(|G|/|H|\). Let’s see if this works for another group.

Example 4.2 Consider the subgroup \(H = \{e, r_1, r_2\}\) of \(D_3\). Here the distinct left cosets of \(H\) in \(D_3\) are: \(eH = r_1H = r_2H = \{ e, r_1, r_2 \}\) \(s_1H = s_2H = s_3H = \{ s_1, s_2, s_3 \}.\)

Which looks like:

\[\begin{array}{c|cc} \circ & H & s_1H\\ \hline H & H & s_1H \\ s_1H & s_1H & H \end{array}\]

So far, so good. Now let’s try another.

Example 4.3 Consider the subgroup \(H = \{e, s_1\}\) of \(D_3\). Here the distinct left cosets of \(H\) in \(D_3\) are: \[eH = s_1H = \{ e, s_1, \}, r_1H = s_3H = \{ r_1, s_3 \}, r_2H = s_2H = \{ r_2, s_2 \}.\]

This time the Cayley table is not partitioned into coset ‘blocks’. We will see that the process of ‘factoring’ a group \(G\) by one of its subgroups \(H\) can yield useful information about \(G\), but first we need to examine exactly when the procedure works.

4.1 Normal Subgroups and Quotient Groups

Let \(H\) be a subgroup of a group \(G\). Now \(H\) has both left and right cosets and, in general, a left coset \(aH, \; a \in G,\) need not be the same as the right coset \(Ha\). Suppose we try to define a binary operation on left cosets as follows: \[(aH)(bH) = (ab)H. \tag{4.1}\]

It is important to realise that this rule does not necessarily define a binary operation on the left cosets of \(H\) in \(G\). This is because it is defined in terms of elements \(a\) and \(b\) but, in general, the cosets will have other representatives and we need to show that it does not matter which representative we choose (this is the same as the situation involving the representatives of the equivalence classes modulo some integer \(m\)).

First recall that \(a' \in aH\) and \(b' \in bH\) if and only if \(aH = a'H\) and \(bH = b'H\) (cosets are either distinct or equal).

Second, recall that a binary operation gives a unique result. In particular for Equation 4.1 to define a binary operation, we must have that \[ (aH)(bH) = (a'H)(b'H) \] or, equivalently, \[ (ab)H = (a'b')H. \]

Consider the case where \(G=D_3\) and \(H= \{e,r_1,r_2\}\). We have: \[\begin{align*} H &= r_1 H = r_2 H = \{e,r_1,r_2\} \\ s_1H &= s_2H = s_3H = \{s_1,s_2,s_3\}. \end{align*}\]

It then follows, for example, that: \[\begin{align*} (r_1H)(s_1H)&=(r_1s_1)H = s_3 H \\ (r_1H)(s_2H)&=(r_1s_2)H = s_1 H \\ (r_2H)(s_3H)&=(r_2s_3)H = s_1 H \end{align*}\] In particular, we see that: \[ (r_1H)(s_1H) = (r_1H)(s_2H) = (r_2H)(s_3H) \]

Note that, although we have changed the representative of the coset, this does not alter the outcome.

But, suppose \(G = D_3\), \(H = \{e,s_1\}\). The distinct left cosets are, \[\{e,s_1\} = eH = s_1H; \ \{r_1, s_3\} = r_1H = s_3H; \ \{r_2,s_2\} = r_2H = s_2H.\]

Then, \[\begin{align*} (eH)(r_1H) &= (er_1H) = r_1H\\ (s_1H)(s_3H) &= (s_1s_3H) = r_2H\\ \end{align*}\] but \(r_1H \ne r_2H\). In this example, Equation Equation 4.1 does not define a binary operation.

It is important to realise what this does (and does not) mean.

With \(G = D_3\) and \(H \in \{e,r_1,r_2\}\), then \[\begin{align*} s_1H &= \{ s_1 e,s_1r_1, s_1r_2 \} = \{s_1,s_2,s_3 \} \\ Hs_1 &= \{es_1, r_1s_1,r_2s_2\} = \{s_1,s_2,s_3\}. \end{align*}\] So \(s_1H = Hs_1\) even though \(s_1 r_1 \ne r_1 s_1\) for instance. However, for \(H = \{e,s_1\}\) \[\begin{align*} r_1H &= \{ r_1 e,r_1s_1\} = \{r_1,s_3 \} \\ Hr_1 &= \{er_1, s_1r_1\} = \{r_1,s_2\}. \end{align*}\] so \(r_1H \ne Hr_1\).`

Definition 4.1 (Normal Subgroup) A subgroup \(H\) of a group \(G\) is normal if its left and right cosets coincide, that is if \(gH = Hg \; \forall \, g \in G\).

We shall often use the notation \(H \norm G\) to denote that \(H\) is a normal subgroup of \(G\). The next Lemma tells us that \((*)\) defines a binary operation provided that, for all elements of the group, the left and right cosets coincide for a given subgroup. The Lemma is, in fact, an ‘if and only if’ statement; we prove only the forward implication below as the converse does not add anything to what we need.

Lemma 4.1 Let \(H\) be a subgroup of a group \(G\). If \(gH = Hg\) for all \(g \in G\), then left coset multiplication is well-defined by \((aH)(bH) = (ab)H\).

(It does not matter which representative we choose in any given coset - in this context that is what we mean by well-defined.)

Proof.

Let \(a,a',b,b' \in G\) satisfy \(aH = a'H\) and \(bH = b'H\). We show that \[(aH)(bH)= abH = a'b'H = (a'H)(b'H).\]

There are elements \(h,k \in H\) such that \(a'h = a\) and \(b'k = b\). Therefore \[a'b'H = ahbkH = ahbH\] since \(kH = H\) as \(k \in H\).

Observe that \(hb\) is an element of the right coset \(Hb\). Since \(Hb = bH\), there is an element \(j \in H\) such that \(hb = bj\) and so \[a'b'H = ahbH = abjH = abH\] as required,

Note

If \(G\) is abelian then every subgroup is normal!

As a convention we will often use \(N\) to denote a normal subgroup. The following result is the one towards which we have been building.

Theorem 4.1 Let \(N\) be a normal subgroup of a group \(G\). Then the cosets of \(N\) form a group \(G/N\) under the binary operation \((aN)(bN) = (ab)N\).

Proof.

We note that the set \(G/N\) is non-empty as it contains the coset \(eN = N\).

Since \(N\) is a normal subgroup of \(G\) then by Lemma 4.1 the binary operation is closed.

For associativity, let \(a,b,c \in G\), then \[\begin{align*} (aN bN)(cN) &= abN cN \\ &= (ab)cN \\ &= a(bc)N \\ &= aN(bcN) \\ &=aN(bNcN) \end{align*}\] The identity element if \(eN\).

lastly, the inverse of \(aN\) is \(a^{-1}N\).

The group \(G/N\) in the above theorem is called the quotient or factor group of \(G\) with \(N\). The following Lemma gives the standard way of showing that a subgroup is normal.

Lemma 4.2 Let \(G\) be a group and \(H\) a subgroup of \(G\). Then \(H\) is normal in \(G\) if and only if for all \(g \in G\) \[gH \ginv \subseteq H,\] where \(gH \ginv = \{ gh\ginv \where h \in H \}\).

Proof.

First suppose that \(H\) is a normal subgroup of \(G\). Let \(g \in G\) and \(h \in H\). Since \(gH = Hg\) there is an element \(k \in H\) such that \(gh = kg\). Post-multiplying by \(g^{-1}\) now yields: \[ gh\ginv = kg\ginv = ke = k.\] Therefore \(gh\ginv \in H\). Since \(h \in H\) was arbitrary, we conclude that \(gH\ginv \subseteq H\).

Now suppose that \(gH\ginv \subseteq H\) for all \(g \in G\). Let \(g \in G\). Let \(gh \in gH\) be arbitrary. Then, \[gh = ghe = gh(g^{-1}g) = (ghg^{-1})g = kg\] where \(k \in H\) is equal to \(ghg^{-1}\) since by assumption \(gHg^{-1}\subseteq H\). Therefore we conclude that \(gh \in Hg\). Since \(h \in H\) was arbitrary we see that \(gH \subseteq Hg\)

Similarly, given \(k \in H\), then observe that \(kg = gg^{-1}kg\). Since \(g^{-1}\in G\), then by assumption, \(g^{-1}Hg \subseteq H\) and so there is \(h \in H\) such that \(g^{-1}kg = h\). Therefore, \[kg = g(g^{-1}kg) = gh \in gH.\] We conclude that \(Hg \subseteq gH\) and so \(Hg = gH\).

Since \(g \in G\) was arbitrary, we conclude that \(gH = Hg\) for all \(g \in G\) and \(H\) is a normal subgroup.

For any non-abelian group an important normal subgroup of \(G\) is the centre of G. This consists of all the elements that commute with every other element of the group.

Definition 4.2 (Centre) Let \(G\) be a group and \(z, g \in G\), then the centre of \(G\) is denoted and defined by \[Z(G) = \{ z \in G \; | \; zg = gz, \; \forall \, g \in G \}.\]

Obviously, a group is abelian if and only if \(Z(G) = G\), so this concept is only of interest for non-abelian groups. That the centre forms a normal subgroup is the content of the next theorem.

Theorem 4.2 The centre of a group \(G\) forms a normal subgroup of \(G\).

Proof.

First observe that \(Z(G)\) if it is a subgroup must be a normal subgroup. This follows since for \(z \in Z(G)\) and \(g \in G\) \[ gz\ginv = g \ginv z = z. \] Therefore \(gZ(G)\ginv = Z(G)\) for all \(g \in G\).

We now show that \(Z(G)\) is a subgroup.

First notice that \(e \in Z(G)\) since \(eg = ge = g\) for all \(g \in G\).

Let \(y,z \in Z(G)\) and consider \(yz\). Let \(g \in G\), then \[ yzg = y(zg) = y(gz) = (yg)z = (gy)z = gyz.\] Therefore, \(yz \in Z(G)\).

Lastly consider \((zg^{-1})^{-1}\), we have \[ gz^{-1}=(zg^{-1})^{-1} = (g^{-1}z)^{-1} = z^{-1}g \] and so \(gz^{-1}= zg^{-1}\) and \(z^{-1} \in Z(G)\).

Example 4.4  

We consider the centre of various group we have already encountered:

• \(Z(D_3) = \{e\}\);
• \(Z(D_4) = \{e, r_2\}\);
• \(Z(D_5) = \{e\}\);
• \(Z(D_6) = \{e, r_3\}\);
• \(Z(A_4) = \{e\}\);
• \(Z(\Z_2 \times D_4) = \{(0,e),(1,e),(0,r_2),(1,r_2)\}\).

4.2 Further Examples of Normal Subgroups

Example 4.5 For any group \(G\), the trivial subgroup \(\{e\}\) is normal and, since \(\{e \}\) contains only one element, every coset of \(\{e\}\) contains only one element. It is therefore easy to see that \[G/\{e\} \cong G\] since all we have done is to rename \(g \in G\) by the coset \(g\{e\} = \{g\}\).

Example 4.6 At the other extreme, we have that for any group \(G\), \(G\) itself is a normal subgroup of \(G\) and \(G/G\) is the trivial group — the group of order 1.

These two extremes of quotient groups are of little importance because they tell us nothing new about the structure of \(G\). We now consider an example where \(H\) is a subgroup of \(G\) and has half the size of \(G\). Consider \(a \in G \backslash H\) (so \(a \notin H\)); then \(aH\) must be all of \(G \backslash H\) (because \(H\) and \(aH\) are disjoint and of equal size and \(a \in aH\)) and, similarly, \(Ha\) must be all of \(G \backslash H\), so \(aH = Ha\). Thus it follows that any subgroup that is half the size of the group must be normal in that group and \(|G| = 2 |N|\).

Example 4.7 Since \(|S_n| = 2|A_n|\) we have that \(A_n\) is a normal subgroup of \(S_n\) and \(S_n/A_n\) has order 2. If \(\sigma\) is any odd permutation, for example \((1 \; 2) \in S_n\), then \(\sigma \in S_n \backslash A_n\), the elements of \(S_n/A_n\) are \(\{A_n, \sigma A_n\}\) and its Cayley table is:

\[\begin{array}{c|cc} & A_n & \sigma A_n\\ \hline A_{n} & A_{n} & \sigma A_{n} \\ \sigma A_{n} & \sigma A_{n} & A_{n} \end{array}\]

Example 4.8 Lagrange’s Theorem states that if \(H\) is a subgroup of a finite group \(G\), then the order of \(H\) divides the order of \(G\). In the Abstract Algebra module we stated that the converse is false; we are now in a position to prove this by showing that \(A_4\), which has order 12, contains no subgroup of order 6.

Proof. Suppose \(A_4\) had a subgroup \(N\) of order \(6\), then \(N\) must be a normal subgroup as it has only \(2\) left cosets \(N\) and \(\sigma N\) for any \(\sigma \in A_4 \backslash N\). Let \(\sigma\) be any element of \(A_4\) that is not an element of \(N\), we can write down the Cayley table of the quotient group \(A_4/N\),

\[\begin{array}{c|cc} & N & \sigma N\\ \hline N & N & \sigma N \\ \sigma N & \sigma N & N \end{array}\]

Notice that \[NN = (eN)(eN) = N = (\sigma \sigma N) = \sigma N \sigma N.\] From this it follows that for any \(\sigma \in A_4\), \((\sigma^2 )N =N\). This means that \(\sigma^2 \in N\) for any \(\sigma \in A_4\). However for any three cycle \((a \; b \; c)\), it square, \((a \; c \; b)\) is also a three cycle. This means that \(N\) contains all three cycles. However \(A_4\) has \(8\) three cycles and so \(|N| > 8\) which is a contradiction.

Example 4.9 Classify the quotient group \((\Z_4 \times \Z_6)/ \langle (0, 1) \rangle\) according to the Fundamental Theorem of Finitely Generated Abelian Groups.

Let \[N = \gen{(0,1)} = \{(0,0),(0,1),(0,2),(0,3)(0,4),(0,5)\}.\] Since \(\Z_4 \times \Z_6 = 24\), then \((\Z_4 \times \Z_6)/N\) has order \(4\). By the Fundamental Theorem of Finitely Generated Abelian Groups, \((\Z_4 \times \Z_6)/N\) is isomorphic to \(\Z_4\) or \(\Z_2 \times \Z_2\).

Notice that each of \((1,0), (2,0)\) and \((3,0)\) give rise to distinct cosets since \((a,0)((b,0))^{-1} = (ab^{-1},0) \notin N\) for any \(a,b \in Z_4\) with \(a\) and \(b\) not both \(0\). Therefore the quotient group \((\Z_4 \times \Z_6)/N\) is generated by \((1,0)N\) since \((1,0)^2N =((1,0)N)^2 = (2,0)N\), \((1,0)^3 N =((1,0)N)^3 = (3,0)N\) and \((1,0)^4N = ((1,0)N)^4 = N\).

The above example is a special case of the following theorem.

Theorem 4.3 Let \(G = H \times K\) be the direct product of groups \(H\) and \(K\). Then \[\overline{K} = \{ (e, k) \; | \; k \in K \}\] is a normal subgroup of \(G\) isomorphic to \(K\) and \(G / \overline{K} \cong H\).

Proof.

It is easily verified that \(\overline{K}\) is a subgroup of \(G\) and the map \((e,k) \mapsto k\) from \(\overline{K} \to K\) is an isomorphism.

Let \((h,k) \in G\) be arbitrary, then: \[\begin{align*} (h,k)\overline{K} &= \{ (h,k)(e,k'): k' \in K \}\\ &= \{ (h,kk'): k' \in K \}\\ &= \{ (h,k''): k'' \in K \}. \end{align*}\] Therefore, \((h,k)\overline{K} = \{(h,k'): k' \in K\}\). Similarly, \[\begin{align*} \overline{K}(h,k) &= \{ (h,k'k): k' \in K \}\\ &= \{ (h,k''): k'' \in K \}. \end{align*}\] Therefore, \[(h,k)\overline{K} =\overline{K}(h,k)\] and \(H\) is a normal subgroup.

Observe that the order of \(G/\overline{K}\) is equal to the order of \(H\). Moreover, for all \(h_1,h_2 \in H\) with \(h_1 \ne h_2\), \[(h_1,e) K = \{ (h_1,k): k \in K \} \ne \{(h_2, k): k \in K\} = (h_2,e) K.\] Therefore, the left cosets of \(\overline{K}\) (the elements of \(G/\overline{K}\)) are \[\{ (h,e)\overline{K} : h \in H\}.\]

It now follows that the map \(\phi: G/\overline{K} \to H\) by \((h,e)\overline{K} \mapsto h\) is an isomorphism.

The following two examples provide further twists on looking at subgroups of the right-hand factor of the direct product, in the sense that in Example 4.10 and Example 4.9, the cyclic subgroup makes no contribution to the left-hand factor whereas that will not be the case in Example 4.11.

Example 4.10 Classify the quotient group \((\Z_4 \times \Z_6)/ \langle (0, 2) \rangle\) according to the Fundamental Theorem of Finite Abelian Groups.

\(\begin{array}{c|c|c} &\text{Order}&\hspace{1cm}\theta: G \rightarrow \Z_4 \times \Z_2\\ \hline \vphantom{\D \frac{a}{b}} \quad \kbar = \langle (0, 2) \rangle = \{ \hspace {6cm} \}&1&\\ \hline \vphantom{\D \frac{a}{b}}(1, 0) + \langle (0, 2) \rangle = \{ \hspace {6cm} \}&\\ \hline \vphantom{\D \frac{a}{b}}(0, 1) + \langle (0, 2) \rangle = \{ \hspace {6cm} \}&\\ \hline \vphantom{\D \frac{a}{b}}(2, 0) + \langle (0, 2) \rangle = \{ \hspace {6cm} \}&\\ \hline \vphantom{\D \frac{a}{b}}(3, 0) + \langle (0, 2) \rangle = \{ \hspace {6cm} \}&\\ \hline \vphantom{\D \frac{a}{b}}(1, 1) + \langle (0, 2) \rangle = \{ \hspace {6cm} \}&\\ \hline \vphantom{\D \frac{a}{b}}(2, 1) + \langle (0, 2) \rangle = \{ \hspace {6cm} \}&\\ \hline \vphantom{\D \frac{a}{b}}(3, 1) + \langle (0, 2) \rangle = \{ \hspace {6cm} \}&\\ \end{array}\)

The group \(G = (\Z_4 \times \Z_6)/ \langle (0, 2) \rangle\) has order \(24/3 = 8\) and isomorphic to \(\Z_4 \times \Z_2\). Below we give two isomorphisms. The red terms give the image of the corresponding coset under the first isomorphism and the blue terms under the second. Notice that once we specify the images of the generators \((1,0) + \overline{K}\) and \((0,2) + \overline{K}\) the remaining terms are determined. To work out the order of an element, we are effectively answering the question ‘’What is the smallest positive integer power of that particular representative of the given coset that lands in the subgroup that is the identity?’’

\(\begin{array}{c|c|c} &\text{Order}&\hspace{1cm}\theta: G \rightarrow \Z_4 \times \Z_2\\ \hline \vphantom{\D \frac{a}{b}} \quad \kbar = \langle (0, 2) \rangle = \{ (0,0),(0,2),(0,4) \}&1& \quad {\color{red}{(0,0)}}\qquad \qquad \qquad {\color{blue}{(0,0)}}\\ \hline \vphantom{\D \frac{a}{b}}(1, 0) + \langle (0, 2) \rangle = \{ (1,0), (1,2),(1,4) \}& 4 & \quad {\color{red}{(1,0)}} \qquad \qquad \qquad {\color{blue}{(1,1)}} \\ \hline \vphantom{\D \frac{a}{b}}(0, 1) + \langle (0, 2) \rangle = \{ (0,1),(0,3),(0,5) \}& 2 & \quad {\color{red}{(0,1)}} \qquad \qquad \qquad {\color{blue}{(0,1)}} \\ \hline \vphantom{\D \frac{a}{b}}(2, 0) + \langle (0, 2) \rangle = \{ (2,0),(2,2),(2,4) \}& 2 & \quad {\color{red}{(2,0)}} \qquad \qquad \qquad {\color{blue}{(2,0)}}\\ \hline \vphantom{\D \frac{a}{b}}(3, 0) + \langle (0, 2) \rangle = \{(3,0),(3,2),(3,4) \}& 4 & \quad {\color{red}{(3,0)}} \qquad \qquad \qquad {\color{blue}{(3,1)}}\\ \hline \vphantom{\D \frac{a}{b}}(1, 1) + \langle (0, 2) \rangle = \{ (1,1),(1,3),(1,5)\}& 4 & \quad {\color{red}{(1,1)}} \qquad \qquad \qquad {\color{blue}{(1,0)}}\\ \hline \vphantom{\D \frac{a}{b}}(2, 1) + \langle (0, 2) \rangle = \{ (2,1),(2,3),(2,5) \}& 2 & \quad {\color{red}{(2,1)}} \qquad \qquad \qquad {\color{blue}{(2,1)}} \\ \hline \vphantom{\D \frac{a}{b}}(3, 1) + \langle (0, 2) \rangle = \{(3,1),(3,3),(3,5) \}& 4 & \quad {\color{red}{(3,1)}} \qquad \qquad \qquad {\color{blue}{(3,0)}} \end{array}\)

Example 4.11 Classify the quotient group \((\Z_4 \times \Z_6)/ \langle (2, 3) \rangle\) according to the Fundamental Theorem of Finite Abelian Groups.

The subgroup \(N = \gen{(2,3)}\) has order two and is equal to \(\{(0,0),(2,3)\}\). Therefore taking \(G = \Z_4 \times \Z_6\), \(G/N\) has order \(12\). Two elements \((a,b)\) and \((c,d)\) belong to the same coset of \(G\) if and only if \((a-c,b-d) \in N\) That is \(a = c\) and \(b=d\) or \(a\) and \(c\) differ by \(2\) and \(b\) and \(d\) differ by \(3\). This gives left cosets: \[\begin{align*} &(0,0)N,(0,1)N,(0,2)N,(0,3)N,(0,4)N,(0,5)N,\\ &(1,0)N,(1,1)N,(1,2)N,(1,3)N,(1,4)N,(1,5)N. \end{align*}\]

Since \(|G/N| =12\), then by the Fundamental Theorem of Finite Abelian Groups \(G/N\) is either isomorphic to \(\Z_{2} \times \Z_{6}\) or to \(\Z_{12}\). However, we observe that \(G/N\) contains an element of order \(4\), \((1,0)N\). Since \(((1,0)N)^2 = (2,0)N = (0,3)N\) and \(((0,3)N)^2 = (0,0)N = N\). Therefore, \(G/N\) is isomorphic to \(\Z_{12}\) and is in fact generated by \((1,2)N\).

4.3 The Correspondence Theorem

In general, a finite quotient group \(G/N\) will be smaller than \(G\) and we may hope that it has a simpler structure. It would be nice if knowing the structure of \(G/N\) helped us to understand the structure of \(G\). Our main result, in this regard, is the following theorem which we state for finite groups (the statement is also true for infinite groups, but the proof is longer and we shall only use the finite case).

Theorem 4.4 (The Correspondence Theorem) Let \(N\) be a normal subgroup of a finite group \(G\). Then \(H\) is a subgroup of \(G\) containing \(N\) if and only if \(H/N\) is a subgroup of \(G/N\), and every subgroup of \(G/N\) has this form for some \(H\) in \(G\).

What this theorem says is that if \(N\) is a normal subgroup of \(G\) then, given a subgroup of \(G\) containing \(N\), we can produce a unique subgroup of the quotient group \(G/N\) and, conversely, given a subgroup of the quotient group we can produce a corresponding subgroup of \(G\) that contains \(N\).

We leave the proof until we have studied an illustrative example of how the Correspondence Theorem works.

First consider the direct product group \(\Z_2 \times \Z_2\).

\[\begin{array}{c|cccc} \oplus_{2}& (0,0) & (0,1) & (1,0) & (1,1)\\ \hline (0,0) & (0,0) & (0,1) & (1,0) & (1,1) \\ (0,1) & (0,1) &(0,0) & (1,1) & (1,0) \\ (1,0) & (1,0) & (1,1) & (0,0) & (0,1) \\ (1,1) & (1,1) & (1,0) & (0,1) & (0,0) \end{array}\] Subgroups of order 2 are: \[\begin{align*} H_1 &= \{(0,0), (0,1)\} \\ H_2 &= \{(0,0), (1,0)\} \\ H_3 &= \{(0,0), (1,1)\} \end{align*}\]

Excluding the trivial group \(\{(0,0)\}\) and \(\Z_2 \times \Z_2\) itself, these are all the subgroups of \(\Z_2 \times \Z_2\).

Now we consider the group \(G=D_4\) and its normal subgroup \(N = \{e, r_2\}\).

The order of \(D_4\) is \(8\) and the order of \(N\) is \(2\), therefore the order of \(D_4/N\) is \(4\). Thus, \(D_4/N\) is isomorphic to \(\Z_4\) or \(\Z_2 \times \Z_2\).

The subgroups of \(D_4/N\) are:

\[\begin{align*} X_1 &= \{N,r_1N\}; N \cup r_1 N = \{e,r_1, r_2, r_3\} \le D_4 \\ X_2 &= \{N, s_1N\}; N \cup s_1N = \{ e, r_2, s_1, s_3\} \le D_4 \\ X_1 &= \{N,s_2N\}; N \cup s_2 N = \{e,r_2, s_2, s_4\} \le D_4. \end{align*}\]

Now we prove the theorem.

Theorem 4.5 (The Correspondence Theorem) Let \(N\) be a normal subgroup of a finite group \(G\). Then \(H\) is a subgroup of \(G\) containing \(N\) if and only if \(H/N\) is a subgroup of \(G/N\), and every subgroup of \(G/N\) has this form for some \(H\) in \(G\).

Proof (proof of the correspondence theorem).

Let \(H\) be a subgroup of \(G\) that contains \(N\). Notice that since \(N\) is normal in \(G\), then \(N\) is normal in \(H\). This follows as since \(hN = Nh\) for all \(h \in G\), then \(hN = Nh\) for all \(h \in H \le G\). It therefore follows that \(H/N\) is a group, however \(H/N \subset G/N\) and so \(H/N\) is a subgroup of \(G/N\).

Now suppose \(X\) is a subgroup of \(G/N\). Let \(\{N, x_1N, x_2N, \ldots, x_rN\} \subseteq G/N\) be the distinct elements of \(X\). Define \[H := \bigcup_{1 \le i \le r} x_i N = \{ x_i n : 1 \le i \le r, n \in N\}.\] If \(H\) is a subgroup of \(G\), then it is a subgroup of \(G\) that contains \(N\) and \(H/N\) is precisely the subgroup \(X\) of \(G/N\). Thus we need only show that \(H\) is a subgroup of \(G\). Clearly \(H\) is non-empty since it contains \(N\). Let \(x,z \in H\). There are \(x_i, x_j\), \(1 \le i,j \le r\) and \(n_1, n_2 \in N\) such that \(x_i n_1 = x\) and \(x_jn_2 = y\). Therefore \[xy = x_i n_1 x_j n_2 = x_i e n_1 x_j n_2 = x_i x_j x_j^{-1}n_1 x_j n_2 = x_i x_j (x_j^{-1}n_1 x_j) n_2.\] Since \(N\) is a normal subgroup of \(G\), \((x_j^{-1}n_1 x_j) \in N\) and so there is an \(n\in N\) such that \((x_j^{-1}n_1 x_j)n_2 = n\). Thus, \[xy = x_ix_j n.\] Lastly, as \(X\) is a subgroup of \(G/N\), then there is a \(k\) between \(1\) and \(r\) such that \[(x_iN)(x_jN) = (x_ix_j)N = x_kN.\] Thus, there is an element \(m \in N\) such that \(xy = x_ix_jn = x_km\) and so \(xy \in H\). This means that \(H\) is closed under products.

One can similarly show that \(H\) is closed under inverses, however, as \(H/N\) is finite, it is enough to show closure under products.

Note

One can actually demonstrate the uniqueness of the group \(H\) above. For if \(H\) and \(H'\) satisfy \(H/N = H'/N\), then for \(x \in H\), there is an \(x' \in H'\) such that \(xN = x'N\). In particular, there is an \(m \in N\) such that \(x = x'm\) and so, as \(N \le H'\), \(x \in H'\). This demonstrates that \(H \subseteq H'\); swapping the order of \(H\) and \(H'\) above, we also have that \(H' \subseteq H\). Thus, \(H = H'\) as required.

Example 4.12  

Let \(G\) be the group \(D_4\). Let \(N\) be the normal subgroup of \(D_4\), \(N = \{e, r_2\}\). Since \(|D_4| = 8\), then \(D_4/N\) has order \(4\). It follows that \(D4/N\) is either \(\Z_4\) or \(\Z_2 \times \Z_2\). However, we observe that every element of \(D_4/N\) has order \(2\). Indeed we need only check the element \(r_1N\): \[ (r_1N)^2 = r_1^2 N = r_2 N = N.\] We conclude that \(D_4/N\) is in fact \(\Z_2 \times \Z_2\).

Now \(D_4/N\) has three subgroups of order two each of which corresponds to a subgroup of order \(4\) in \(D_4\):

\[ \begin{array}{c|c} \text{Subgroup of } N & \text{Corresponding subgroup of } D_4 \\ \hline \{N, r_1N\} & \{e,r_2,r_1,r_3\} \\ \{N, s_1N\} & \{e,r_2,s_1,s_3\} \\ \{N, s_2N\} & \{e,r_2,s_2,s_4\} \end{array} \] Thus we have all thee subgroups of \(G\) of order \(4\) each containing \(N\) and corresponding to exactly one of the subgroups of order \(2\) in \(D_4/N\).

Example 4.13 Demonstrate the proof of the Correspondence Theorem where the quotient group is \[(\Z_4 \times \Z_6)/\langle(0, 2)\rangle.\]

We demonstrated in Example 4.10 that \((\Z_4 \times \Z_6)/\langle(0, 2)\rangle\) is isomorphic to \(\Z_4 \times \Z_2\).

Non-trivial (that is not equal to the subgroups \(\{(0,0)\}\) and \(\Z_4 \times \Z_2\)) subgroups of \(\Z_4 \times \Z_2\) have order \(2\) and \(4\). There are \(3\) distinct subgroups of order \(4\) \[ \gen{(1,0)}, \gen{((1,1))}, \text{ and, } \gen{(2,0), (0,1)}. \] Each element of order \(2\) of \(\Z_4 \times \Z_2\) generates a unique subgroup of order \(2\), there are \(3\) of them: \[ \gen{(2,0)}, \gen{((2,1))}, \text{ and, } \gen{((0,1)}. \]

Using Example 4.10, and setting \(N = \gen{(0,2)}\), we can write down the correspondence between the subgroups of \((\Z_4 \times \Z_6)/\langle(0, 2)\rangle\) and \(\Z_4 \times \Z_6\)

\[ \begin{array}{c|c} \text{Subgroups of order 4 } & \text{Corresponding subgroup of } \Z_4 \times \Z_6 \\ \hline \{N, (1,0)N,(2,0)N, (3,0)N\} & \{e,(0,2),(0,4),(1,0),(1,2),(1,4),\\ &(2,0),(2,2),(2,4),(3,0),(3,2),(3,4)\} \\ & \\ \{N, (1,1)N,(2,0)N,(3,1)N\} & \{e,(0,2),(0,4),(1,1),(1,3),(1,5),\\ & (2,0),(2,2),(2,4),(3,1),(3,3),(3,5)\} \\ & \\ \{N, (0,1)N, (2,1)N, (2,0)N\} & \{e,(0,2),(0,4),(0,1),(0,3),(0,5),\\ &(2,1),(2,3),(2,5),(2,0),(2,2),(2,4)\} \end{array} \]

\[ \begin{array}{c|c} \text{Subgroups of order 2 } & \text{Corresponding subgroup of } \Z_4 \times \Z_6 \\ \hline \{N, (0,1)N\} & \{e,(0,2),(0,4),(0,1),(0,3),(0,5)\} \\ \{N, (2,0)N\} & \{e,(0,2),(0,4),(2,0),(2,2),(2,4)\} \\ \{N,(2,1)N\} & \{e,(0,2),(0,4),(2,1),(2,3),(2,5)\} \end{array} \]

Example 4.14  

Let \(G\) be a group of order \(44\). Suppose \(G\) has a normal subgroup \(N\) of order \(11\), so that \(G/N\) is a group of order \(4\). Thus \(G/N\) is isomorphic either to \(\Z_4\) or \(\Z_2 \times \Z_2\). Now both \(\Z_4\) and \(\Z_2 \times \Z_2\) have subgroups of order \(2\) and so \(G\) has a subgroup \(H\) containing \(N\) such that \(H/N\) has order \(2\). This means that \(H\) has order \(22\) and so must itself be a normal subgroup of \(G\). Thus \(H\) is a normal subgroup of \(G\) containing \(N\).

4.4 Normalisers

We now consider a question which may seem artificial, but will turn out to be very important.

Given a subgroup, \(H\), of a group \(G\), what is the largest subgroup of \(G\) containing \(H\) as a normal subgroup?

Consider the group \(H = \{e,r_2\} \norm D_4\). The largest subgroup of \(D_4\) in which \(H\) is normal, is \(D_4\) itself.

Now \(\{e,s_1\}\) is not a normal subgroup of \(D_4\), but \(\{e, r_2, s_1,s_3\}\) is a subgroup of \(D_4\) in which \(\{e, s_1\}\) is normal. Notice that the elements \(g \in D_4\) such that \(g\{e,s_1\} = \{e,s_1\}g\) are precisely \(e,r_2, s_1\) and \(s_3\).

We had better make sure that this is a sensible question! If \(H\) is a normal subgroup of \(G\), then the answer is \(G\). If \(H\) is not normal in \(G\) can we be certain that there is a subgroup of \(G\) that contains \(H\) as a normal subgroup? Yes! Any group is a normal subgroup of itself so \(H\) is a subgroup of \(G\) containing \(H\) as a normal subgroup. Having established the existence of such a subgroup it makes sense to ask for the largest such subgroup (except that there may be more than one). Let \(L\) be a largest subgroup of \(G\) containing \(H\) as a normal subgroup. It is certainly the case that any element \(l\in L\) has the property that \(lH=Hl\). It would be nice if every element \(g\in G\) with the property that \(gH=Hg\) formed a subgroup of \(G\), then this subgroup would certainly answer our question. We first define this set, then prove that it is a subgroup.

Definition 4.3 (Normaliser) Let \(H\) be a subgroup of a group \(G\). The normaliser of \(H\) in \(G\) is denoted and defined by \[N(H)=\{g\in G| gH=Hg\}.\]

Theorem 4.6 Let \(H\) be a subgroup of a group \(G\), then the normaliser of \(H\) in \(G\) forms a subgroup of \(G\).

Proof.

We prove this both directly, in the normal way, and indirectly using a group action.

• Direct proof: Clearly \(e \in N(H)\) since \(eH = H = He\). Let \(g_1, g_2 \in N(H)\). Then \[g_1g_2H = g_1 (g_2H) = g_1(Hg_2) = (g_1H)g_2 = H(g_1g_2)\] and so \(g_1g_2 \in N(H)\). Let \(g \in N(H)\) and consider \(g h_1\) for \(h_1 \in H\). There is and \(h_2 \in H\) such that \(g h_1 = h_2 g\). Taking inverses of both sides \[h_1^{-1}g^{-1} = (gh_1)^{-1} = (h_2g)^{-1} = g^{-1}h_2^{-1}.\] We conclude that \(g^{-1}H = Hg^{-1}\).

• Indirect proof:
  We define an action \(\ast\) of \(G\) on the set of all of the subsets of \(G\). In terms of the definition of a group action (see Definition 3.1) this set is the set denoted \(X\) and we need to demonstrate that all three conditions hold. Clearly for \(S \subseteq G\), \(gS \ginv\) is again a subset of \(G\). Thus \(g \ast S \in X\) for all \(S \in X\) and \(g \in G\). Moreover,

  • \(e \ast S = e S e^{-1} = e S e = S\) for all \(S \in X\);
  • \(g_1 \ast (g_2 \ast S = g_1 \ast (g_2 S g_2^{-1}) = g_1g_2 S g_2^{-1}g_1^{-1} = g_1g_2 S (g_1g_2)^{-1} = (g_1g_2) \ast S\). So \(\ast\) is indeed an action of \(G\) on the set of all subsets of \(G\). We know that the stabiliser of an element of \(X\) under the action is a subgroup of \(G\). However, the stabiliser of a subgroup under this particular action, is just the normaliser of the subgroup yielding the result.

4.5 Problem Sheet 4

For Week 9; covers Chapter 4.

---

Question 4.1

Classify each of the following groups according to the Fundamental Theorem of Finite Abelian Groups and in each case give an isomorphism from the stated group to the group given as your answer.

1. \((\Z_3 \times \Z_6) / \langle (1, 0) \rangle\).
2. \((\Z_4 \times \Z_8) / \langle (1, 2) \rangle\).
3. \((\Z_4 \times \Z_8) / \langle (2, 2) \rangle\).

Solution 4.1

1. \((\Z_3 \times \Z_6) / \langle (1, 0) \rangle\) is an abelian group of order 6 (as \(|\, (\Z_3 \times \Z_6) \, | = 3 \times 6 = 18\) and \(| \, \langle (1, 0) \rangle \, | = 3\)) so it must be isomorphic to \(\Z_6\) since there is only one group of this order that is abelian (up to isomorphism). But \(\Z_6\) is not in the form given by the FTFAG as 6 is not the power of a prime. So the given group is isomorphic to \(\Z_2 \times \Z_3\) (which is, of course, isomorphic to \(\Z_2 \times \Z_3\)). You should check that this follows from Theorem 5.3. An isomorphism from \((\Z_3 \times \Z_6) / \langle (1, 0) \rangle\) to \(\Z_6\) is easy (just map \((x, y) + \langle (1, 0) \rangle\) to \(y\) ), but we need a mapping from \((\Z_3 \times \Z_6) / \langle (1, 0) \rangle\) to \(\Z_2 \times \Z_3\). Since both groups are cyclic we just map a generator in one group to a generator in another, for example: \[\begin{equation*} (0, 1) + \langle (1, 0) \rangle \mapsto (1, 1). \end{equation*}\] It then follows that \[\begin{align*} (0, 2) + \langle (1, 0) \rangle &\mapsto (0, 2)\\ (0, 3) + \langle (1, 0) \rangle &\mapsto (1, 0)\\ (0, 4) + \langle (1, 0) \rangle &\mapsto (0, 1)\\ (0, 5) + \langle (1, 0) \rangle &\mapsto (1, 2)\\ (0, 6) + \langle (1, 0) \rangle &\mapsto (0, 0) \qquad \text{as expected!} \end{align*}\]
2. This one is not so obvious, so we need to kick it around a bit. \[\langle (1, 2) \rangle = \{(0, 0), (1, 2), (2, 4), (3, 6) \} = H.\] Now we can see that \((0, 1) + H\) has order 8 (because the first multiple of \((0, 1)\) in \(H\) is \((0, 0)\)) and since \((\Z_4 \times \Z_8) / \langle (1, 2) \rangle\) has order 8 it must be isomorphic to \(\Z_8\). An isomorphism is given by \[(x, y) + H \mapsto y.\] Note that this is in the correct form for the FTFAG as 8 is a power of a prime.
3. Again, this is not obvious, so let \[H = \langle (2, 2) \rangle = \{ (0, 0), (2, 2), (0, 4), (2, 6) \}.\] Hence, \((\Z_4 \times \Z_8) / \langle (2, 2) \rangle\) has order 8. This time \((0, 1) + H\) has order 4 (make sure you understand why). This rules out \(\Z_2 \times \Z_2 \times \Z_2\), but still leaves \(\Z_8\) and \(\Z_2 \times \Z_4\) as possibilities. We could list the order of each of the elements, or look for elements of order 8, or for more than one element of order 2 (because cyclic groups of even order have exactly one element of order 2). You should beware of falling into the following trap. It may be tempting to say that, since \((0, 2)+H\) and \((2, 0)+H\) both have order 2, the group is isomorphic to \(\Z_2 \times \Z_4\). But, \((0, 2)+H\) and \((2, 0)+H\) are both the same coset (list the elements if you cannot see this directly)!! However, \((1, 1)+H\) and \((0, 2)+H\) are distinct elements of order 2 so the group is indeed isomorphic to \(\Z_2 \times \Z_4\). The following is an isomorphism \[\begin{align*} H \mapsto (0, 0) &\qquad (1, 1) + H \mapsto (1, 0)\\ (0, 1) + H \mapsto (0, 1) &\qquad (1, 2) + H \mapsto (1, 1)\\ (0, 2) + H \mapsto (0, 2) &\qquad (1, 3) + H \mapsto (1, 2)\\ (0, 3) + H \mapsto (0, 3) &\qquad (1, 4) + H \mapsto (1, 3) \end{align*}\]

Question 4.2

Let \(H\) be a subgroup of a group \(G\).

1. Show that, for all \(g \in G\), the set \[gH\ginv = \{ gh\ginv \, | \, h \in H \}\] forms a subgroup of \(G\).
2. Deduce that if \(G\) has exactly one subgroup, \(H\), of a given finite order, then \(H\) is a normal subgroup of \(G\).

Solution 4.2

1. Clearly \(gH\ginv \neq \emptyset\) and \(gH\ginv \subseteq G\) so we just need to show that the set \(gH\ginv\) is closed under products and inverses. Let \(g_1, g_2 \in gH\ginv\). So, there exist \(h_1, h_2 \in H\) such that \(g_1 = gh_1\ginv\) and \(g_2 = gh_2\ginv\). Then \[g_1g_2 = (gh_1\ginv)(gh_2\ginv) = gh_1h_2\ginv \in gH\ginv\] (because \(h_1h_2 \in H\) as \(H\) is a subgroup). Also, \[g_1^{-1} = (gh_1\ginv)^{-1} = (\ginv )^{-1} h_1^{-1} \ginv = gh_1^{-1} \ginv \in gH\ginv\] (because \(h_1^{-1} \in H\) as \(H\) is a subgroup).
2. If \(H\) is the only subgroup of order \(|H|\) then, for all \(g \in G\), \(gH\ginv = H\) (think about it…). That is, \(H\) is normal in \(G\).

Question 4.3

Let \(H\) be a subgroup of a group \(G\). Prove that the centre of \(G\) is a normal subgroup of the normaliser of \(H\). [Hint: all you really need to show is that \(Z(G) \subseteq N(H)\) and the rest is ‘obvious’.]

Solution 4.3

First recall the definitions. The centre of \(G\) is denoted and defined by \[Z(G) = \{ z \in G \, | \, zg = gz \; \forall \, g \in G \}.\] The of \(H\) in \(G\) is denoted and defined by \[N(H) = \{ g \in G \, | \, gH = Hg \}.\] Now we know that \(N(H)\) is a subgroup of \(G\) and we also know that \(Z(G)\) is a normal subgroup of \(G\). So, if we can show that \(Z(G) \subseteq N(H)\), then it follows that \(Z(G)\) is a normal subgroup of \(N(H)\). Now, let \(x \in Z(G)\). Then, \[\begin{align*} & xg = gx \qquad \forall \, g \in G\\ \imp & xh = hx \qquad \forall \, h \in H \qquad \text{(as $h \in G$ also)}\\ \imp & xH = Hx\\ \imp & x \in N(H). \end{align*}\] Hence \(Z(G) \subseteq N(H)\), as required. (Note that \(Z(G)\) must be normal in \(N(H)\) else it would not be normal in \(G\).)

Question 4.4

Let \(n \ge 3\). Show that the centre of \(S_n\) is trivial. What is the centre of \(S_2\)?

Solution 4.4

Suppose \(g \in S_n\) is a non-trivial element of \(Z(S_n)\). There is a point \(x \in \{1,2,\ldots, n\}\) such that \(g(x) = y \ne x\). Let \(z \in \{1,2,\ldots, n\}\backslash\{x,y\}\). Notice that \(z\) exists since \(n \ge 3\).

Consider \(g(x\ y \ z)g^{-1}\). Since \(g \in Z(S_n)\), then \(g(x \ y \ z)g^{-1} = (x \ y \ z)\). We compute \[g(x \ y \ z)g^{-1} = (g(x) \ g(y) \ g(z)) = (y \ g(y) \ g(z)) = (x \ y \ z).\] It follows that \(g(y) = z\) and \(g(z) = x\).

Now consider the two cycle \((x \ y)\). We must also have \(g(x \ y)g^{-1} = (x \ y)\). However, from above, \(g(x \ y)g^{-1} = (y z) \ne (x \ y)\) (since \(z \ne x\)). This yields the desired contradiction. We conclude that \(Z(S_n)\) contains only the identity permutation.

The group \(S_2\) is isomorphic to \(Z_2\) and so is abelian and equal to its own centre.

Question 4.5

Let \(N\) be a normal subgroup of a finite group \(G\), and \(H\) be a further subgroup of \(G\) (which need not be normal in \(G\)). Show that the set \(NH\), where \[NH = \{ nh \, | \, n \in N, h \in H \}\] forms a subgroup of G.

Solution 4.5

Clearly \(e \in NH\) so \(NH \neq \emptyset\). Since \(G\) and, hence, the set \(NH\) is finite we need only check the closure of \(NH\) under products. Let \(n_1, h_1, n_2, h_2 \in NH\). Then, \[\begin{align*} (n_1h_1)(n_2h_2) &= n_1(h_1n_2)h_2 \qquad \text{(associativity)}\\ &= n_1(n_3h_1)h_2 \qquad \text{(for some} \; n_3 \in N \; \text{because} \; h_1N = Nh_1)\\ &= (n_1n_3)(h_1h_2) \; \in NH \qquad \text{(as} \; n_1n_3 \in N \; \text{and} \; h_1h_2 \in H). \end{align*}\] Hence \(NH\) is a subgroup of \(G\).

Question 4.6

In \(D_4\), let \(e\) and \(r_2\) denote the identity and \(\pi\) radians rotation respectively. Given that \(N = \{e, r_2\}\) is a normal subgroup of \(D_4\), give an example of a surjective homomorphism \[\phi : D_4 \maps D_4 / N\] and classify \(D_4 / N\) according to the Fundamental Theorem of Finite Abelian Groups.

Solution 4.6

Note that \(|D_4| = 8\) and \(|N| = 2\), so \(|D_4 / N| = 4\). Hence it follows (from the Fundamental Theorem) that either \(D_4/N \cong \Z_4\) or \(D_4/N \cong \Z_2 \times \Z_2\). The elements of \(D_4/N\) are: \[\begin{align*} N = r_2N &= \{e, r_2\}\\ r_1N = r_3N &= \{r_1, r_3 \}\\ s_1N = s_3N &= \{s_1, s_3 \}\\ s_2N = s_4N &= \{s_2, s_4 \}. \end{align*}\] Clearly \((s_1N)^2 = (s_2N)^2 = N\) and \((r_1N)^2 = r_2N = N\), so all non-identity elements have order 2. Hence \(D_4/N \cong \Z_2 \times \Z_2\). A surjective homomorphism \(\phi : D_4 \maps D_4 / N\) is \[\begin{eqnarray*} \phi(e) = \phi(r_2) &= N\\ \phi(r_1) = \phi(r_3) &= r_1N\\ \phi(s_1) = \phi(s_3) &= s_1N\\ \phi(s_2) = \phi(s_4) &= s_2N. \end{eqnarray*}\]

Question 4.7

Consider the quarternion group \(Q_8\) with elements \(\{\pm 1, \pm i, \pm j, \pm k\}\) and multiplication defined by the rules: \[\begin{eqnarray*} (-1)^2 = 1,& \\ i^2 = j^2 = k^2= -1,& \end{eqnarray*}\] and \[ij=k = -(ji), jk =i=-(kj), ki= j = -(ik).\]

The group \(N = \{1, -1\}\) is a normal subgroup of \(Q_8\). Find elements \(g_1, g_2, \ldots, g_i \in Q_8\) such that \(\{g_1N, g_2N \ldots, g_iN\}\) are the distinct left cosets of \(N\) in \(G\).

Classify the group \(Q_8/N\) according to the Fundamental Theorem of finite abelian groups.

Solution 4.7

The distinct left cosets are given by \(N, iN, jN, kN\). This follows since for distinct elements \(a,b \in \{i,j,k\}\), \(-ab \in \{\pm i, \pm j, \pm k\}\). Now observe that in \(Q_{8}/N\), the elements \(iN, jN\) and \(kN\) all have period \(2\). We conclude that \(Q_{8}/N\) is isomorphic to \(\Z_2 \times \Z_2\).